I am trying to reproduce the computations of Appendix B of Fractal Concepts on Surface Growth (Barabási & Stanley), about the computation of the critical exponents of KPZ equation using Dynamic Renormalization Group techniques. I have the KPZ equation,

$$\frac{\partial h\left(\vec{x},t\right)}{\partial t}=v\nabla^{2}h+\frac{\lambda}{2}\left(\nabla h\right)^{2}+\eta\left(\vec{x},t\right)$$

where $\eta$ is a Gaussian white noise with correlations $\left\langle \eta\left(\vec{x},t\right)\eta\left(\vec{x}',t'\right)\right\rangle =2D\delta\left(\vec{x}-\vec{x}'\right)\delta\left(t-t'\right)$.

Then, I do a perturbative expansion and solve the integrals to renormalize $\nu$. I have no problem with that. I get the result indicated on the book,


where the integral is divergent for $d\leq d_c = 2$ as $q \rightarrow 0$. In section B.3, the book tells how to renormalize the integral in $dq$, and it gives two steps:

1) The equivalent of coarse graining in real space, which to integrate in a momentum shell $\Lambda /b \leq q \leq \Lambda$, leaving the integral $0 \leq q \leq \Lambda / b$ untouched. We will use also $b=e^\ell\simeq(1+\ell)$.

2) A rescaling, which in space is $x\rightarrow bx$ and in momentum is $q \rightarrow q/b$.

However I don't understand how this is done in the divergent integral. What you can see in the book is that they compute

$$\int_{\Lambda/b} ^{\Lambda}q^{d-3}dq \simeq \int_{\Lambda (1-\ell)} ^{\Lambda}q^{d-3}dq \simeq \ell \Lambda^{d-2}$$

where they let $\Lambda=1$ without loss of generality. Then they say that this result is the one we get when we are with long wavelenghts (slow modes) only,


And then they apply the rescaling as $\tilde{\nu} =b^{z-2}\nu^< \simeq \nu^< [1+\ell(z-2)]$. Then the book substitutes the expression of $\nu^<$ and operates to order $\mathcal{O}(\ell)$ to find the flow equations for the parameter.

Under my point of view, what it is doing is to integrate the fast modes and put this result inside the expression of $\nu^<$, which are the slow modes, and forgets about the integral between 0 and $\Lambda/b$.

So my question is: I don't understand why we integrate over the fast modes and then substitute directly in the expression of $\nu$. More precisely, I don't know what happened to the integral between 0 and $\Lambda/b$. What am I missing here?.

I tried to see why this is done in this way. I tried to separe the integral in slow and fast modes:

$$\int_{0} ^{\Lambda}q^{d-3}dq= \int_{0} ^{\Lambda/b}q^{d-3}dq + \int_{\Lambda/b} ^{\Lambda}q^{d-3}dq$$

Then the second integral can be done and it is a constant. As I see, when I rescale $q\rightarrow q/b$, the first integral has the limit $\Lambda/b \rightarrow \Lambda /b^2$. In addition to that, the constant will depend also on $\Lambda$, so it will be also rescaled -unless we use the trick to put $\Lambda = 1$ as they do, so I am really not sure on how this works.

So, how this separation into slow and fast modes happen? Any explanation and/or useful sources are welcome.

UPDATE: I added a bounty for getting more attention into this question. In addition to that, I want to point out that I've read that in fact all this come from the beginning, so I have to split the field $h(\vec{k},\omega)=h^<(\vec{k},\omega)+h^>(\vec{k},\omega)$ and then average only over the fast modes $h^>(\vec{k},\omega)$. This will give the expression for $\nu ^<$ I am trying to find, after doing to one-loop integral, that will be only in the momentum shell. However I couldn't get a lot of detail on how to do this process. Thank you!

  • $\begingroup$ What do you know about Wilson renormalization group ? If the answer is nothing, start with that, as your question will be answered in any basic level RG lecture. If you do know it, could you explain where is the problem ? $\endgroup$
    – Adam
    May 23, 2017 at 11:05
  • $\begingroup$ I have read about RG and I believe I understand the basics. The problem is that when I do the integrals up to first order, I find $\bar{\nu}=\nu(1-I)$, where $I$ is an integral in $0<q<\Lambda$. However, in the book, they simply solve $\nu ^< = \nu(1-I^>)$, where $I^>$ is in $\Lambda/b<q<\Lambda$. I understand that the coarse graining means integrating out the high $q$ momenta, but, where did $I^<$, the integral in $0<q<\Lambda/b$, go? I have tried to find something detailed in the bibliography and I found nothing. $\endgroup$ May 23, 2017 at 11:14
  • $\begingroup$ I think I understand your problem, see my answer, and let me know if it is indeed what was troubling you. $\endgroup$
    – Adam
    May 23, 2017 at 12:28

1 Answer 1


The OP's problem seems to come from a slight misunderstanding of the RG procedure. Not having the OP's reference available, I will comment on the standard $\phi^4$ theory, which should be sufficient to answer the problem, the action of which is $$S=\int_x\left[ \frac{(\nabla \phi)^2}{2}+\frac{r}{2}\phi^2+\frac{u}{4!}\phi^4\right].$$

The coupling $\nu$ seems to be equivalent to the mass term $r$, so I will only discuss the renormalization of this term.

The idea of Wilson RG is to integrate out high-momenta modes $\phi_>$ while leaving untouched the slow modes $\phi_<$. The partition function is written as $$Z=\int D\phi_< \int D\phi_> e^{-S[\phi_< ,\phi_>]}.$$

Then $S[\phi_< ,\phi_>]$ is split into $S_<[\phi_<]$ which involves only slow modes, and $S_{int}[\phi_< ,\phi_>]$ which involves the interaction between the slow and the fast modes. After integrating out the fast modes, of momentum $\Lambda/b <q<\Lambda$, one obtains a new action for the slow modes, $$Z=\int D\phi_< e^{-S_{<,b}[\phi_<]},$$ with $$S_{<,b}[\phi_<]=S_<[\phi_<]-\log\left[\int D\phi_> e^{-S_{int}[\phi_< ,\phi_>]}\right].$$ Notice that by definition, the correction to $S_<[\phi_<]$ coming from the log can only involve loops with integrals over momenta $q\in[\Lambda/b ,\Lambda]$.

Performing a loop expansion of the log term, we obtain a quadratic term in the slow modes of the form ($\alpha$ is some constant which we could figure out) $$\alpha\int_xu\int_{\Lambda/b <q<\Lambda}\langle\phi_>(q)\phi_>(-q)\rangle_{>,0}\phi_<(x)^2\equiv\int_x\delta r_b\phi_<^2(x),$$ which means that the quadratic term of $S_{<,b}[\phi_<]$ is thus $\int_x r_b\phi_<^2(x),$ with $$r_b=r+\delta r_b =r+\alpha\int_xu\int_{\Lambda/b <q<\Lambda}q^{d-3},$$ assuming $\langle\phi_>(q)\phi_>(-q)\rangle_{>,0}\propto q^{-2}$.

We thus see that the renormalized mass term (before rescaling) involves only integrals over high momenta (which is the point of Wilson procedure).

  • $\begingroup$ Thanks for the detailed answer. However, the KPZ equation is a model for a non-equilibrium process, so I cannot define a partition function. Still I could translate my stochastic PDE to a field theory and define an action, but I would have to add an auxiliar field. Any clue your method with the non-equilibrium process? $\endgroup$ May 23, 2017 at 14:22
  • $\begingroup$ Yes, I had in mind a field theory for KPZ, which works (up to technical details) the same way. I guess their method is equivalent to that. $\endgroup$
    – Adam
    May 23, 2017 at 14:36
  • $\begingroup$ Ok, I've found the book. What they are doing here is a dirty way of setting up a RG procedure (that is, they know what to do because they know RG, and cook up some rules to find the correct result). In principle, one could also do that for standard QFT... If you don't want to just accept their rules, set up a RG procedure from the equivalent FT for KPZ, it will be much more satisfying. $\endgroup$
    – Adam
    May 23, 2017 at 14:45

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